If `f(x) =|(1, x, x+1),(2x, x(x-1), x(x+1)), (3x(x-1), 2(x-1)(x-2), x(x+1)(x-1))|`
then what is `f(-1) +f(0) +f(1)` equal to ?

Options

A. 0
B. 1
C. 100
D. `-100`

Answer

To solve the problem, we need to evaluate the determinant given by the function \( f(x) \) and find \( f(-1) + f(0) + f(1) \).

The function is defined as:
\[ f(x) = \begin{vmatrix} 1 & x & x+1 \\ 2x & x(x-1) & x(x+1) \\ 3x(x-1) & 2(x-1)(x-2) & x(x+1)(x-1) \end{vmatrix} \]

### Step 1: Evaluate the determinant

We can simplify the determinant using properties of determinants. One useful operation is to modify one of the columns by subtracting a linear combination of other columns.

Let's apply the operation \( C_3 \rightarrow C_3 - C_1 + C_2 \):
\[ C_3 = (x+1) - 1 + x = 0 \]
This means the third column becomes zero after the operation.

### Step 2: Recognize the determinant value

When any column (or row) of a determinant is entirely zero, the value of the determinant is zero. Therefore:
\[ f(x) = 0 \]

### Step 3: Calculate \( f(-1) + f(0) + f(1) \)

Since we have established that \( f(x) = 0 \) for all \( x \):
\[ f(-1) = 0, \quad f(0) = 0, \quad f(1) = 0 \]
Thus:
\[ f(-1) + f(0) + f(1) = 0 + 0 + 0 = 0 \]

### Final Answer
\[ f(-1) + f(0) + f(1) = 0 \]